Cơ học lượng tử 2.2

1" I. Spinors II. The matrices and eigenspinors of S x and S y 2" II. Spinors This is an example of using the matrix formulation of Quantum Mechanics "Matrix Mechanics" with 2-component i and f states i Recall that a particle can have 2 kinds of angular momentum -spin angular momentum and orbital angular momentum i Recall that spin (a kind of angular momentum) can have components, in particular a z-component i Recall that z-component of spin can have only 2 values: S z = up and S z = down also called + and − (no 0 ) 3" i Recall that a particle's m quantum number concerns the z-component of its angular momentum If the particle has no orbital angular momentum (m  = 0) Then m=m s only. i Recall that the general operator that represents the measurement of m is J z So in general J z Ψ = m Ψ if =0, this is m spin Recall on (?) we showed that quantum # j (like  but including spin) can take values given by integer 2 like L z but generalized to measure m spin AND m  orbital angular momentum and spin angular momentum 4" I. Spinors (continued) II. The matrices and eigenspinors of S x and S y 5" Also m J max = + j and m J min =-j And (m J max − m J min ) = integer The way to satisfy all of this for a 2-state system is for j= 1 2 m J max = + 1 2 m J min = - 1 2 No othe m J values allowed. So if =0, so j = spin only then we have m spin max = + 1 2 m spin min = - 1 2 Make a matrix to reflect J z when j = spin only: Ψ f S z Ψ i = Ψ f m spin  Ψ i = m spin  Ψ f Ψ i = m spin  δ if S z = m spin final + − + + 1 2  0 − 0 − 1 2  ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ m spin initial δ if Call it "S z " 6" Recap what we know: + and − exist They are eigenfunctions of the spin measurement, S z Their eigenvalues are + 1 2  and - 1 2  We can summarize this information as: S z + =  2 + and S z − = -  2 − The matrix representation for S z is +  2 0 0 -  2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ What are the matrix representations for + and − ? To answer this we need to solve +  2 0 0 -  2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ u + z v + z ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = +  2 u + z v + z ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ and +  2 0 0 -  2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ u − z v − z ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = −  2 u − z v − z ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ S z S z + − Dirac notation stuff Matrix notation stuff 7" The solutions are alternative symbol S, S z = + 1 2 , + 1 2 + = u + z v + z ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 1 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = χ + z eigenfunctions of S z = + 1 2 , - 1 2 − = u − z v − z ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 0 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = χ − z *Any 2 component vector is called a spinor *These two-component column vectors which are the eigenfunctions of S z in the m s basis are called the eigenfunctions of S z We could also write for example a matrix S x to describe the measurement of the x-component of spin That matrix would have different eigenfunctions χ + z and χ + z , the "eigenspinors of S x " Any pair of spinors: (1) are orthogonal: χ + χ + = 0 (2) are normalized: χ + χ + = χ − χ − = 1 (3) form a basis in "spin space": any state of arbitrary spin χ can be represented by a linear combination of them χ =a χ + + b χ − coefficients also called the S z basis 8" II. The matrices and eigenspinors of S x and S y Recall in P 491 we showed that [L x , L y ] = iL z (and cyclic x → y → z) and we postulated that [J x , J y ] = iJ z (and cyclic x → y → z) Now postulate that S x , S y , S z are related in the same way: [S x ,S y ] = iS z (and cyclic x → y → z) Also recall from P 491 the definition of the general angular momentum raising and lowering operators: J + ≡ J x + iJ y J − ≡ J x − iJ y Since these are general, they raise or lower both orbital angular momentum and spin angular momentum When L=0, they act only on S, so we could call them in that case: S + ≡ S x + iS y S − ≡ S x − iS y L S 9" To make the matrix for S x (in the m s or the S z basis) we need the matrices for S + and S − in the m s basis: Recall: J + j,m j =  j( j + 1) − m j (m j + 1) j,m j + 1 In general j=+s and m j = m  + m s Suppose =m  = 0 Then j=s and m j =m s Then S + s,m s =  s(s + 1) − m s (m s + 1) s,m s + 1 But s= 1 2 only =  1 2 ( 3 2 ) − m s (m s + 1) s,m s + 1 So s,m s ′ S + s,m s =  3 4 − m s (m s + 1) δ m s ′ ,m s +1 Make the matrices: J − j,m j =  j( j + 1) − m j (m j − 1) j,m j − 1 S − s,m s =  s(s + 1) − m s (m s − 1) s,m s − 1 =  1 2 ( 3 2 ) − m s (m s − 1) s,m s − 1 s,m s ′ S − s,m s =  3 4 − m s (m s − 1) δ m s ′ ,m s −1 10" m s : + 1 2 - 1 2 S + = m s ′ : + 1 2 − 1 2 0  3 4 − (− 1 2 )(− 1 2 + 1) 0 0 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 0  0 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Combine these to get S x = S + + S − 2 = 1 2 0  0 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + 0 0  0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 1 2 0   0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =  2 0 1 1 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ S y = S + − S − 2i = 1 2i 0  0 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 0 0  0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = −i 2 0  − 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =  2i 0 −i +i 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Now find their eigenspinors eigenvectors in the m s basis m s : + 1 2 - 1 2 S − = m s ′ : + 1 2 − 1 2 0 0  3 4 − ( 1 2 )( 1 2 −1) 0 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 0 0  0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . solve S + χ ± x = λ ± χ ± x 0  2  2 0 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ u x v x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = λ u x v x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − λ  2  2 − λ ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ u x v x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0 − λ  2  2 − λ = 0 λ 2 −  2 4 = 0 λ = ±  2 To get the. 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = −i 2 0  − 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =  2i 0 −i +i 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Now find their eigenspinors eigenvectors in the m s basis m s : + 1 2 - 1 2 S − = m s ′ : + 1 2 − 1 2 0 0  3 4 − ( 1 2 )( 1 2 −1). =  1 2 ( 3 2 ) − m s (m s − 1) s,m s − 1 s,m s ′ S − s,m s =  3 4 − m s (m s − 1) δ m s ′ ,m s −1 10" m s : + 1 2 - 1 2 S + = m s ′ : + 1 2 − 1 2 0  3 4 − (− 1 2 )(− 1 2 + 1) 0